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Limit rational function without L'H

  1. Mar 15, 2013 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations
    [itex]lim_{x->0+} \frac{\sqrt{x}}{\sqrt{sinx}}[/itex]

    3. The attempt at a solution
    i've tried l'hopital's and it is just endless cycle.
  2. jcsd
  3. Mar 15, 2013 #2


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    It's the same as sqrt(x/sin(x)). You know the limit of x/sin(x), right?
  4. Mar 15, 2013 #3


    Staff: Mentor

    Sometimes, L'Hopital's Rule is not the way to go. Under the right conditions, you can switch the order of the limit operation and the function in the limit.
    ## \lim f(g(x)) = f(\lim g(x))##

    Also, as long as all quantities are positive,
    $$ \frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$$
  5. Mar 15, 2013 #4
    yes, but it is of the form 0/0.
  6. Mar 15, 2013 #5


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    That doesn't mean you HAVE to use l'Hopital. You know the limit of x/sin(x), use l'Hopital on that. Then take the square root. Use that the square root is continuous for positive arguments.
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